Group Theory Basics: Where Can I Learn More?

[Hurkyl]

Just to let everyone know, I probably won't be able to think about this for a while.

 

[Lonewolf]

Oh dear. lol. I'm quitting work this Friday, so I'll have time to give the topic the attention it needs. Be glad to hear your input when you return, Hurkyl. Always nice to have two people explaining something.

 

[marcus]

Originally posted by Lonewolf
Oh dear. lol. I'm quitting work this Friday, so I'll have time to give the topic the attention it needs. Be glad to hear your input when you return, Hurkyl. Always nice to have two people explaining something.

I agree----two or MORE.
Several times Hurkyl assigned "homework problems" or steps in the exposition for the reader to prove and Lonewolf or I would solve stuff or fill in the gaps which is a good kind of dialog. And
Lethe may have gotten in there too.

So Lonewolf, we should consider what kind of holding operation we want during Hurkyl's absence

I could for instance do a chapter on something entirely different but related to the representation theory of groups/algebras

or I could keep on with the basic (Hall's book) expo of matrix
groups and algebras---pretending to be Hurkyl, in effect.

Also LETHE! any comment? ideas of where thread could go?
want to be the substitute schoolteacher while Hurkyl's away?

 

[Lonewolf]

Do as you see fit. I'm kinda struggling to get to grips with representations, but then again, I haven't given it much attention as of yet. I'd quite like to see "something entirely different". I'm largely using this as a preview of years to come, but I also want to get something solid out of it, if you know what I mean.

 

[marcus]

Baez: Oz and the Wizard story about Representations

Originally posted by Lonewolf
Do as you see fit. I'm kinda struggling to get to grips with representations, ...

Have you seen this story about Representations by John Baez?
-------------"spr" quote-----------------
"Well," said the Wiz, putting down his alembic and scratching his
head, "I'll try to explain it with a minimum of math..."

Oz smiled. "Good. As Feynman said, we don't really understand
anything unless we could explain it to our mother"

At this the Wiz bristled. "Speak for yourself. *My* mother knew
quite a bit about this stuff... she was a high priestess of the
Pythagorean Order... that's how I was born knowing tensor calculus."

"Really?" said Oz, not sure whether the Wiz was having him on.
"But enough of my personal life," said the Wizard impatiently.
"You'll probably disappear any minute now, so let me give you a
quick crash course on this stuff you're wondering about."
He thought a moment, running his fingers through his beard, and
then launched into an explanation:

"You know about various physical quantities: mass, energy, velocity, momentum, angular momentum, and so on. And you probably know that it's useful to keep track of their dimensions - in terms of length, time and mass, for example - at least that's how they teach you dimensional analysis in grade school. But: why is this so useful?"

"Well," said Oz confidently, "it keeps you from making mistakes: it
keeps you from adding apples and oranges and getting potatoes, so to speak. Sometimes you can even guess what the answer has to be just by remembering that the units work out right."

"Exactly!" said the Wiz. "Now, when you did this stuff, did you
ever realize you were doing group representation theory?"
Oz's eyes bulged. "Group... representation... theory? Err, no.
That's wizard talk, that is! I know nothing of that."

"Well, you were. Group representation theory is just the study
of how quantities change when apply various sorts of transformations to them. For example, if I tell you how long something is in rods, and ask you to tell me what its length is in feet, what do you?"

"Why, multiply by sixteen and a half, of course," replied Oz,
instantly losing his terrified demeanor. "Every schoolchild knows that!" "Right; that's how it transforms. But now say I tell you an *area*and ask you to convert it from units where length is measured in rods to units where length is measured in feet. What do you do then?"

"Multiply by sixteen and a half *squared*, of course, area has
units of length squared." "Right: area transforms in a different representation," agreed the Wiz.
"Hey, wait a minute!" said Oz. "You're trying to trick me into
thinking I understand this representation theory business, when I
actually don't!"The Wiz smiled. "No: I'm trying to trick you into realizing you actually DO understand it better than you think! When we change units of length, various physical quantities transform in various ways. We can actually imagine expanding all distances by a factor of 2 and seeing how various quantities change: some would stay the same (like times), some would double (like distances), others would quadruple (like areas), while other would be divided by 8 (like densities). Each of these ways of transforming is called a "representation" - in this case, a representation of the group of dilations."

"Dilations?" asked Oz, his pupils widening in terror.
"Yes, that's just a fancy wizard-word for stretching. Anyway,
when you're solving a physics problem, you know it doesn't make
sense to add a distance and an area, because they transform
differently under dilations, so even if your calculation *happened*
to work out correctly in units of feet, it wouldn't in units of rods."
"Right," said Oz. "But what does this have to do with vectors,
and pseudovectors, and bivectors, and ..."
"Well," said the Wizard, "Just as it's handy to keep track of
how quantities transform under dilations, as a bookkeeping device to keep from making silly mistakes, it's also handy to keep track of how they transform under *rotations*, and other sorts of transformations. Only here the options are more varied. For example, we have quantities like mass or energy, that don't change at all under rotations... we
call these SCALARS."

--------to be continued---------

 

[marcus]

Baez story about Representations, ctd

-----exerpt from sci.physics.research, "spr"-----

Oz nodded, then thought more deeply and got confused. "What do you mean, they don't change under rotations?"

The Wiz glared. "I mean just what I say! See this weight?" With
a wave of his wand, a bang and a puff of smoke, an enormous weight labelled 50 TONS appeared on the floor. Oz held his hand over his eyes and squinted, leaning forward.
"Yes... but you have to realize, everything keeps fading in and out, over there!"

"Well, suppose we rotate it." With another wave of the wand the Wiz conjured up an enormous greenish troll, who grabbed the weight and turned it a bit, and then stared dumbly at it, drool oozing from between his half-open lips. "What's it's mass now?"

Oz rolled his eyes at the enormous expenditure of magic being
wasted on such a simple point. "Why, exactly what it did before!"
"Right!" said the Wiz. He snapped his fingers, and the weight
and the troll disappeared. "Mass doesn't change at all under rotations, so we call it a scalar! On the other hand, something like velocity does! We can measure the velocity of a bullet in some Cartesian coordinate system and get 3 numbers: the x, y and z components."

He pulled out a rusty old flintlock from one of the cabinets and fired it out the window. The glass shattered; the bullet left a trail of smoke, magically labelled by 3 numbers. "If we rotate the experiment and do it again, we get different numbers." He turned...

"Hey, don't point that thing at me!" yelped Oz.

"Okay, hopefully you get the point," said the Wiz. "It's a nuisance
having these windows repaired, after all. The point is, we know a
specific rule for how the numbers change when we do a rotation. Or at least *I* do. Do *you* remember it?"

"Umm, err..." said Oz. "I think maybe I sort of vaguely do, though
not quite. You take the numbers, line them up to form a column, and then you multiply them by a matrix... a square box of numbers... you do this by moving your left finger across the box, while moving your right finger down the column, multiplying the numbers and adding them up as you go... it's rather mysterious, come to think of it!"

"Yes, it's actually rather profound," said the Wiz, smiling. "But
for now, my only point is that for any rotation you... or at least
*I* ... can work out a 3 x 3 matrix which tells us how a velocity
transforms under that rotation. Anything that transforms according to this rule, we call a VECTOR. For example, not only velocity, but also momentum, is a vector."

"Okay," continued the Wiz. "How many other ways are there for
physical quantities to transform under rotations?" Oz thought and thought, but couldn't decide. "In other words," said the Wiz, "How many other REPRESENTATIONS are there of the ROTATION GROUP? This is just wizard-speak for the same
question... I don't expect it to help you just yet... I'm only mentioning it so that when you hear wizards muttering about group representations, you'll have more of a sense of what they're up to."

"Yes," said Oz, "that's helpful already. But - how many ARE there?"

"Lots!" said the Wiz. "But the wonderful thing is, I have a list,
which I keep up here," he said, tapping on his forehead, "of what they all are!"

"Hmm!" said Oz. "Could you, umm, tell me what they all are?" On
second thought, getting a bit scared, he backed off a bit. "Or, at
least some of them?"

"Well, for starters I'll tell you this: every different sort of TENSOR
gives you a different representation of the rotation group. To take the simplest example: the stress tensor."

Oz gulped. "Stress tensor? That's the simplest example? It sounds scary... I always get stressed out when you start talking abstract math, and now you're making me even tenser!"

"It's simple, honest!" said the Wiz. "Take this block of rubber" - with a wave of his hand, one appeared in his palm - "and twist, stretch or squash it however you like." He almost tossed it to Oz, but reconsidered. "Hmm, if you're really in a parallel universe, Oz, that may be risky. I'll do it myself."

He stretched it out and twisted it. "Now, imagine how each tiny piece of this rubber feels stretched, squashed or twisted. We can describe this with numbers, but not with 3 numbers - it takes 6!
In fact, we can arrange them in a 3 x 3 matrix, but it's a symmetric matrix: the entry in the ith row and jth column equals that in the jth row and ith column, so there are only 6 independent entries."

Oz looked puzzled. "Symmetric matrices... symmetric rank-2 tensors -- are those the same thing?"

"Yes," said the Wiz, "for now at least - they transform the same way under rotations, anyway. And that's just the point! You see --"

"Wait! I don't really understand it all yet. Where do we get this
matrix from? What do all the numbers mean?"

"Well," said the Wiz, "I don't really want to get into this now, but
the 3 numbers down the diagonal say how much the rubber is being squashed in the x, y, and z directions... or stretched, if the
number is negative. The other 3 numbers say how much and which way it's being twisted. Hmm. I thought you learned all this stuff in the general relativity tutorial!"

"Well, maybe I did, Sir - I do remember a "stress-energy tensor",
vaguely, but that was a 4 x 4 matrix, and it had to do with pressure and energy density and..."

The Wiz cut him off impatiently. "Yes, that's another aspect of
the same idea. Back then we were doing SPACETIME, so we had 4 dimensions, but right now we're just doing SPACE, to keep things simple... anyway, the details don't matter here: I was just trying to give you another example of a representation of the rotation group. That is, a physical quantity that doesn't transform like a scalar when you rotate it, and doesn't transform like a vector. The stress tensor is basically a batch of 6 numbers - arranged artistically in a matrix - and there is a rule, which I will not tell you now, for how the stress tensor of this piece of rubber transforms when I rotate it."

"Oh!" said Oz, "Please tell me the rule, please do..."

"NO!" thundered the Wiz. "I can sense your time here is dwindling to a close. I only have time for this: by keeping track of how things transform under ROTATIONS, we can avoid foolish mistakes like adding things that transform differently, so it is profitable to CLASSIFY ALL REPRESENTATIONS OF THE ROTATION GROUP - and every mathematical physicist worth his or her salt knows this classification. It basically amounts to listing all possible sorts of TENSORS, of which the scalars and vectors
are the very simplest kinds."

"But," he continued, "this is just the beginning. You can do even better if you also keep track of how things transform under reflections! For example: angular momentum transforms just like a vector under rotations, but differently when we do reflections. Have you ever looked a moving object in a mirror, and wondered precisely how the velocity of the mirror image is related to that of the original object?"

 

[marcus]

Baez introduces spin concept, end of story

-------exerpt of Baez post on "spr"----

"Umm, I can't say as I have, though it must be fairly simple."

The Wiz grew even more impatient. "No? What a stunning lack of
curiosity... anyway, do it sometime!

You will then know how a VECTOR transforms under reflections. Then, compare a *spinning* object to its mirror image, and figure out how their angular momenta are related. The rule is different! So we say that angular momentum is a PSEUDOVECTOR! This means that adding velocity and angular momentum is as bad as adding apples and oranges."

"But I already knew that," said Oz. "Velocity and angular momentum have different units!"

"Yes," the Wiz growled, "but even if they DIDN'T, it would STILL
be bad. If I had time, I could invent an example of quantities
with exactly the same units, but one a vector and the other a
pseudovector. But I don't! Or more precisely, *you* don't have
time. Next: what do you get if you take the dot product of two vectors?"

"A scalar!" replied Oz proudly.

"Right! But what if you take the dot product of a vector and a
pseudovector? Like velocity dotted with angular momentum?"

"Umm," said Oz, guessing wildly, "a PSEUDOSCALAR?"

"Right!" said the Wiz. "A scalar doesn't change under reflections;
a pseudoscalar changes sign under reflections."

Oz scratched his head, trying to work it out.

"Anyway," said the Wiz, "I hope you get the pattern: as we consider more and more sorts of transformations - dilations, rotations, reflections -- our classification of physical quantities according to how they transform becomes every more complicated and subtle... but also more POWERFUL, because we can make finer distinctions. This increase our chances that we can figure out the right answer to a physics problem just by writing down the only possibilities that transform correctly!"

"But what about spinors?" asked Oz.

The Wiz sighed. "Ah yes. Well, when we get to quantum mechanics, we need to replace the rotation group by something bigger and better. The reason is that some physical quantities turn out to change when you apply a 360 degree rotation to them! They only come back to where they were after a 720 degree rotation. They're not scalars, or vectors, or any other sort of tensor: to understand them, we need a group that's LIKE the rotation group, but distinguishes between "no rotation at all"
and "a 360 degree rotation about any axis". This new group has 2 elements for each element of the rotation group, so we call it a DOUBLE COVER of the rotation group."

"Is this, umm, US(2)?" asked Oz.

"You mean SU(2)!" replied the Wiz. "Yes. But the real point is
this: as soon as we discover that the rotation group is not sufficiently sophisticated for quantum mechanics, and we have to replace it by some other group, we had better run down to the math department and ask them to tell us all the REPRESENTATIONS of this group, so we can avoid adding apples to oranges in this brave new world!

And if we do, they'll tell us: `well, you've got your scalar representation, and your vector representation, and all your tensors just as before, but now you've got a whole wad of new ones, the simplest being the so-called SPINOR representation...in fact, you've got one for each spin j = 0, 1/2, 1, ...'"

"Spin!" said Oz. "So that's all it is??"

"Yes, for each spin we have a separate rule saying how things should transform under rotations... or, not really rotations, but these SU(2) transformations, which are a lot like rotations, but a little fancier."

"But where do the Dirac matrices come into it?" asked Oz.

"Well, for any representation of any group, you need a lot of
matrices to describe the rules for how things transform. You
know how it works for vectors... I hope... and all the other cases
are similar, but fancier. For the spinor representation of SU(2),
it helps to have some 2 x 2 matrices called `Pauli matrices' at your
disposal. But that's just about rotations in SPACE. If we switch
to studying SPACETIME, we also have to know how quantities transform under Lorentz transformations! We switch to a bigger group called SL(2,C), work out its representations, and discover that there's still something called the spinor representation... only now, to calculate with it, we need some 4 x 4 matrices called Dirac matrices. That's for "Dirac spinors", actually. There are also "Weyl spinors," which work differently..."

POOF! All of a sudden, Oz disappeared.

The Wiz sighed. "Just when it was getting interesting!"

He turned back to his alembic, picked it up, and started scraping
it off, muttering to himself as he worked. "Well, I hope he learned at least a *little* before taking off like that..."

-----end of post---

I found this at
http://www.lns.cornell.edu/spr/2002-01/msg0038075.html
the Cornell "spr" archive. But it may be other places on the
web as well. I seem to recall Hurkyl giving an address for
this as well as for the Baez tutorial on relativity.

 

[Lonewolf]

Lol, madness. He explains it well though. Just checking out the end? in the link you provided. I read part of it when Hurkyl posted, but wasn't focusing completely on it really, so it didn't sink in too well. Thanks for that.

 

[Lonewolf]

Oh, it seems the end has already been posted. Oops.

 

[Lonewolf]

SO(3) and Orbital Angular Momentum

Imagine an electron in a spherically symmetric attractive potential of some atom’s nucleus. The wavefunction of the electron can, as we all well know, be characterised by 3 quantum numbers n, l, m that are related to the eigenvalues of conserved operators H L² and L_z. However, the energy is 2l+1-fold degenerate, depending only on n and l (for an unpure Coulomb potential. In a pure Coulomb potential, the dependency is only on n). The degeneracy may be explained by the spherical symmetry, independent of θ and φ.

Did I just say ‘spherical symmetry’? Does SO(3) spring to mind? It should. The above explanation for the degeneracy is equivalent to the Schrodinger-Hamiltonian (-h_bar²/2m)∇²+V(r) is invariant under ordinary spatial rotations, which is exactly where SO(3) comes into play. Recall SO(3) is the group of ordinary spatial rotations.

The spherical symmetry of the potential V(r) ensures that the orbital angular momentum L is conserved. Instead of using operators to represent L_x, L_y and L_z, we can use matrices. The L_i generate the (2l+1)x(2l+1) irreducible representations of SO(3). The dimension 2l+1 is associated with the 2l+1 degenerate states.

This degeneracy is removed by introducing a constant magnetic induction field B, which leads to the Zeeman effect. This magnetic interaction adds to the Schrodinger-Hamiltonian a term which is not invariant under SO(3). This leads us to conclude that B is that magic thing, a symmetry breaking term.

 

[Hurkyl]

Phew, it seems I'm not drawn away for as long as I thought I'd be!


I guess we should keep heading towards spin, so I'll introduce a class of representations of SU(2). (And it will turn out that any finite dimensional representation of SU(2) will be isomorphic to one of these)


SU(2) is a group of 2x2 invertible complex matrices which means it acts on 2 dimensional complex vectors... we can reinterpret this action as a transformation on pairs of complex numbers.


Now, let's take the space V_m of m-dimensional homogenous complex polynomials in two variables, y and z. That is, the polynomials of the form:


f(y, z) = Σ_k=0..m a_ky^m-kz^k

Note that f(y, z) is an m+1-dimensional vector space with basis {y^kz^m-k | k in 0 .. m}


I'll leave it as an exercise that, for A in SU(2), the mapping:

(Af)(y, z) -> f(A^-1(y, z))

is a linear invertible mapping, and that A(Bf) = (AB)f, which makes V_m an m+1 dimensional complex representation of SU(2).


I'll leave it as homework problem to prove that this is an irreducible representation (since I don't see right away how to prove it! )

 

[marcus]

Lots of tidbits to check here, among other things

I'll leave it as an exercise that, for A in SU(2), the mapping:

(Af)(y, z) -> f(A^-1(y, z))

is a linear invertible mapping, and that A(Bf) = (AB)f...

a kindergarten question (often the best time) is why did they use
f(A^-1(y, z)) instead of f(A(y, z))?

Would f(A(y, z)) have worked as well? no because then we would have gotten an unwanted reversal: "A(Bf) = (BA)f..."

so let's prove the Hurkyl-stated fact!

let g(y,z) be temporary notation for f(A^-1(y, z))
and let B act on it
(Bg)(y,z) = g(B^-1(y, z)) = -> f(A^-1(B^-1(y, z)))


oops, have to go

 

[Hurkyl]

Yep, the reversal is important!

Note we know another involution of complex matrix rings; the conjugate transpose. We could also have defined the action

(Af)(y, z) = f(A^*(y, z))

For SU(2), of course, this is identical with the representation using the inverse, but for, say, SL(2, C) these would be two different actions (at least superficially)

 

[marcus]

Now I can resume that HW exercise. I got started on the wrong foot earlier. I have to show A(Bf) = (AB)f. Acting on the polynomial f first by B and then by A gives the same as acting by AB.

I will temporarily use the letter g to stand for the polynomial Bf, the intermediate result gotten by first acting with B on f.

[QUOTE
let g(y,z) be temporary notation for f(B^-1(y, z))
and let A act on it
(Ag)(y,z) = g(A^-1(y, z)) = f(B^-1(A^-1(y, z))) = f((B^-1A^-1(y, z))
by associativity
and there's the fact about matrix multiplication that B^-1A^-1 = (AB)^-1, so
QUOTE]

the above is f((AB)^-1(y,z))
which is (AB)f, the result of AB acting on f,
so that shows A(Bf) = (AB)f

 

[marcus]

About these SU(2) matrices, all novices should now know
what the 4 complex numbers are in the general form of
one
Since we seem to be moving on it would ordinarily be time
to review basic knowledge with a midterm, but perhaps since we go by a sort of anti-academic set of rules we will omit this.

given two complex numbers u and v,
with uu* + vv* = 1,
the general SU(2) matrix is

 u  v
-v* u*

And if you invert this by swapping on the main diagonal and negging on the other, in the time-honored way when det = 1, then by the Three Graces and the Seven Muses you get

u* -v
v*  u

and this is also the complex conjugal flip transpose A* thing
so A^-1 = A*

Now the natural naive question to ask is whether any special SU(2) property of the matrix is helping out here. Does the fact that A is SU(2) help make the polynomial f(y,z) map into a new polynomical with is homogeneous with the same total degree?

 

[marcus]

Originally posted by Hurkyl

I'll leave it as homework problem to prove that this is an irreducible representation (since I don't see right away how to prove it! )

The way the group thread works is if Hurkyl assigns homework we do it.

So as a contribution to classifying the irreducible representations of SU(2) I have made two PF footnotes

https://www.physicsforums.com/showthread.php?s=&postid=58530#post58530

which is about complexifying Lie algebras and uniquely extending representations of the algebra to its complexified version, and also this one:

https://www.physicsforums.com/showthread.php?s=&threadid=4671

which is about complexifying the Lie algebra su(2) to get
sl(2, C)

the irreducible reps of sl(2, C) are not dreadfully hard to study
and this will tell us about those of su(2)
which in turn (because reps of the LA correspond to reps of the group)
will tell us about the reps of SU(2)
which will turn out to be on the spaces of homogenous polynomials as we were just now discussing

 

[marcus]

the irred reps of SU(2)

Hurkyl has described an action of SU(2) on the homogeneous polynomials of total degree m-----an m+1 dimensional complex vectorspace.

How may we show that these are irreducible? And that they are ALL the finite dimensional irreps up to isomorphism.

I think Hall page 72-73 sort of does the first part of this, roughly as follows.

Any rep of the group can be lifted to a rep of the Lie algebra su(2), which can be viewed as a representation over the reals and uniquely extended (see footnote) to a rep of sl(2,C) over the complexes.

Call this unique extension [pi]_m
It is a (complex) LA homomorphism from sl(2, C) to the linear operators on the polynomial space V_m

To repeat, V_m is homog. polys with complex coefficients and total degree m, in two variables y,z. as Hurkyl described.

The action with matrix A of sl(2,C) on poly P(y,z) is simply
giving us the new poly P(A^-1(y,z))

[pi]_m(A) sends P(y,z) to P(A^-1(y,z))

NOW WE LOOK AT A BASIS OF sl(2, C), which are TRACE ZERO 2x2 matrices

Following Hall's notation the basis is H, X, and Y where
H = diag(1, -1) = (1,0,0,-1)
X = (0,1,0,0)
Y = (0,0,1,0)
I will use "code" to type these in block form when I edit but here I am just reciting 2x2 matrices left to right and top to bottom as words are written on the page. From the commutation relations we can figure out what the operators [pi]_m(H), [pi]_m(X) , [pi]_m(Y) actually do to the polynomials and show irreducibility directly----no proper invariant subspaces.

Now for the second part, which Hurkyl may be going to prove, we need to show that ANY irreducible rep on ANY finite dimensional V is isomorphic to the polynomial one of the same dimension.

Since we have these three sl(2,C) matrices H,X,Y and know their bracket relations, to study any rep all we need to do is study what it does to these three. Now [pi]_m(H) is an operator on V, so it is actually an (m+1)x(m+1) matrix, assuming V has dimension m+1.
And we can diagonalize it! We can find its eigenvalues and eigenvectors!
This approach parallels Hall pages 76-78 where he proves the theorem that any two irreps of sl(2,C) which have the same dimension are equivalent.
Did not mean to hog the exposition like this! It was really Hurkyl's turn to talk but I got carried away. Sorry.

 

[Hurkyl]

Did not mean to hog the exposition like this! It was really Hurkyl's turn to talk but I got carried away. Sorry.

No problem. Means less work for me. Besides, I think it's helpful to see different styles of explanations for things, so it's good as long as we don't confuse the audience (which, of course, includes ourselves!


I wonder if it might not be good to explore the "geometry" of SU(2) and SL(2, C); we know lots about SO(3), but I don't really have much intuition for those other two groups, and I'm not confident in what I do have. This would be unnecessary for this thread, but might be helpful! But, of course, it might be entirely the case that it would be far easier to understand the geometry after we've gotten through sl(2, C)'s representations instead of before... EDIT: I guess we're nearly there so might as well finish it up.


Incidentally, here's an amusing nonrigorous homework problem! At least I found it amusing when I did it. It seems to require ignoring some "obvious" yet nontrivial technical details, so keep that in mind while you do the problem. (this one's just for fun!)

As we know, so(3) and su(2) are isomorphic three-dimensional real algebras. I demonstrated earlier that so(3) was isomorphic to R³ with the cross product for the bracket, via the basis:

    /  0  0  0 \
[b]i[/b] = |  0  0 -1 |
    \  0  1  0 /

    /  0  0  1 \
[b]j[/b] = |  0  0  0 |
    \ -1  0  0 /

    /  0 -1  0 \
[b]k[/b] = |  1  0  0 |
    \  0  0  0 /

One can check things like [i, j] = i*j. This basis has the nice property that e^it is a t-radian rotation around the R³ vector i, and similarly for j and k.

We know that SO(3) is generated by these three classes of rotations (but is probably nontrival to prove, so don't!)

So what if we take a basis for SU(2)? In particular:

[b]i[/b] = 0.5 /  i  0 \
        \  0 -i /

[b]j[/b] = 0.5 /  0  1 \
        \ -1  0 /

[b]k[/b] = 0.5 /  0  i \
        \  i  0 /

You can check that the bracket again corresponds to the cross product. You can exponentiate these to get elements of SU(2). Recall elements of SU(2) act on su(2) by the adjoint mapping U(X) := UXU^-1. One can then check that e^ti acts as *drumroll*... a rotation of t radians around the vector i! We again assume that SU(2) is generated by these three classes of rotations... but the neat part is that the explicit formula for these exponentials makes it clear why SU(2) is a double cover of SO(3).

 

[MathematicalPhysicist]

proove a theorem in group theory

i need to proove (a^-1)^-1=a with group theory.
the text says the proof should be the same as to the proof to b=a^-1
here this proof:
a.b=e (given)
a^-1.(a.b)=a^-1.e
(a^-1.a).b=a^-1.e
e.b=a^-1.e
b=a^-1

i apply it too:
a^-1.a=e (given)
a^-1.(a^-1.a)=a^-1.e
a^-1.e=a^-1.e
a^-1=a^-1

does this what i neede to come to?
btw I am new to group theory so be gentle (-:

 

[Lonewolf]

Not sure that you've explicitly proven it, though I may be wrong. Try to end up with (a^-1)^-1 = a at the end of your proof.

 

[MathematicalPhysicist]

that's the problem i cant.

 

[Lonewolf]

Try letting b = a^-1, getting a.(b^-1)^-1 = a.a, sub a back in for b^-1, then you're almost there.

 

[marcus]

We used to have a table of how to make greek letters using
font = symbol
I have lost track of where it is or what the symbol font looks like
so I am going to have a look at each key in that font

It is handy because & pi ; gives something that doesn't look very much like pi (in the default) and same for gamma (looks like Y in default) and theta (looks like the number 8 in default)


a in symbol font is a
b in symbol font is b
c in symbol font is c
d in symbol font is d
e in symbol font is e
f in symbol font is f
g in symbol font is g

h in symbol font is h
i in symbol font is i
j in symbol font is j
k in symbol font is k
l in symbol font is l
m in symbol font is m
n in symbol font is n


o in symbol font is o
p in symbol font is p
q in symbol font is q
r in symbol font is r
s in symbol font is s
t in symbol font is t
u in symbol font is u

v in symbol font is v
w in symbol font is w
x in symbol font is x
y in symbol font is y
z in symbol font is z

these are mostly intuitive except that the theta is typed
using the letter q and a couple of things like that. here is
a sample in size = 4 and size = 3 to make the greek letters
more legible

a few size 4:
a in symbol font is a
b in symbol font is b
c in symbol font is c
d in symbol font is d
e in symbol font is e
f in symbol font is f
g in symbol font is g

a sampling of size 3:
h in symbol font is h
i in symbol font is i
j in symbol font is j
k in symbol font is k
l in symbol font is l
m in symbol font is m
n in symbol font is n


ordinary size:
o in symbol font is o
p in symbol font is p
q in symbol font is q
r in symbol font is r
s in symbol font is s
t in symbol font is t
u in symbol font is u

v in symbol font is v
w in symbol font is w
x in symbol font is x
y in symbol font is y
z in symbol font is z

 

[MathematicalPhysicist]

shouldnt this thread should be a sticky one?

 

[Hurkyl]

On my computer, all of those are just roman letters drawn in a fancy way, not greek letters...

 

[pdelong]

errata?

Good (introductory) references are:

M. A. Armstrong. Groups and symmetry. Springer Verlag 1988.

Would you happen to know where one could get an errata listing for this book? I'm currently in the middle of it and I'm stuck on a problem. I'm pretty sure the reason I'm stuck is because of a misprint, but I just want to be sure.

Thanks.

 

[Janitor]

I learned (well, sort of ) my group theory from a moldy-oldie library book written by _____ Hall. It was already old when I was 20, and I imagine it is long out of print.

 

[ahrkron]

On my computer, all of those are just roman letters drawn in a fancy way, not greek letters...

Same here! Mozilla on RedHat Linux. I'll check later on a Windows XP.

 

[ahrkron]

Same thing on XP (Mozilla also).

 

[meteor]

Groups are fantastic things, I'm recently starting to appreciate its importance, before I considered them rather arbitrary constructions
I'm wondering if the group operation is restricted to some binary operations, or instead, any binary operation is valid. Cause I know that for example addition, multiplication, matrix multiplication, composition of functions,... are binary operations that are permitted like group operations, but is this general?, I mean, can any binary operation serve like a group operation?

 

[matt grime]

No, by defintion the binary operation must satisfy certain rules, but by the same token any binary operation satisfying those rules is a group operation. "Serving like a group operation" is a nebulous phrase which could mean anything you chose it to mean.

A group is an axiomatic object, anything satisfying those axioms is a group, end of story. If you want a binary operation that isn't a group operation, there's multiplication on the real line - 0 has no inverse. Or addition on the set of multiples of 2 and 5 - that isn't closed, Addition on the strictly positive real numbers has no identity. For failure of asscoiativity I must be more creative: consider the group defined by this table:

* |a b
--------
a|a b
b|a b


That isn't associative by the failure of the latin square principle.

I forgot to emphasize that your question over looks the fact that there needs to be an underlying set the operation is defined on - matrix multiplication is not a group operation on the set of infinite matrices.

 

[meteor]

Matt,
Yeah, I know that a group has to satisfy certain rules (namely closure, associativity, identity element and existence of the inverse), but my question is: given a group of elements and a binary operation, and then those elements under the binary operation satisfy the rules, then this is considered a group, indepently of the binary operation? I ask this because I do not remember any group where the binary operation is division, or for example the modulus (a mod b), or the Legendre symbol, or many others, but would be nice if such groups could exist
Regards

 

[matt grime]

The binary operation is part of the group structure, it cannot be a group independently of the operation, hell it can't even be a group.

Given a set there are many ways of putting a group structure on it. There are two groups of order 6 that are not isomorphic.

I think you are misuing the word group in that clause 'given a group of elements'. Do you just mean set or group in its proper definition?

Division would not work for associativity reasons.

And those things you cite (legndgre symbol etc) cannot form group operations - one isn't even a binary operation.

 

[Janitor]

As Matt points out, division is not associative. Example: what is 20/10/2?

(20/10)/2 = 1,

20/(10/2) = 4.

 

[matt grime]

I misread what you wrote, they are all binary operations but cannot possibyl be group operations given, firstly you aren't offering a set on which they are to be defined which contradicts the definition that a group is a set with a binary operation, and they fail to define injective maps in that if you fix the first argument there are many (infinitely many) second arguments that will give the same output.

Remember, a group is a set WITH an operation satisfying... you need both.